A function f is defined as follows
`f(x)=x^(p)cos((1)/(x)),xne0" "f(0)=0`
What conditions should be imposed on p so that f may be continuous at x = 0?

Options

A. p = 0
B. `pgt0`
C. `plt0`
D. No value of p

Answer

To determine the conditions on \( p \) for the function \( f(x) = x^p \cos\left(\frac{1}{x}\right) \) (for \( x \neq 0 \)) and \( f(0) = 0 \) to be continuous at \( x = 0 \), we need to check the limit of \( f(x) \) as \( x \) approaches 0.

### Step-by-step Solution:

1. **Define the Function**:
The function is defined as:
\[ f(x) = \begin{cases} x^p \cos\left(\frac{1}{x}\right) & \text{if } x \neq 0 \\ 0 & \text{if } x = 0 \end{cases} \]

2. **Check Continuity at \( x = 0 \)**:
For \( f \) to be continuous at \( x = 0 \), the following must hold:
\[ \lim_{x \to 0} f(x) = f(0) \]
Since \( f(0) = 0 \), we need:
\[ \lim_{x \to 0} f(x) = 0 \]

3. **Evaluate the Limit**:
We need to evaluate:
\[ \lim_{x \to 0} x^p \cos\left(\frac{1}{x}\right) \]
The cosine function oscillates between -1 and 1, hence:
\[ -1 \leq \cos\left(\frac{1}{x}\right) \leq 1 \]
Therefore, we can bound \( f(x) \):
\[ -x^p \leq x^p \cos\left(\frac{1}{x}\right) \leq x^p \]

4. **Apply the Squeeze Theorem**:
As \( x \to 0 \):
- If \( p > 0 \), then \( x^p \to 0 \).
- Thus, by the Squeeze Theorem:
\[ \lim_{x \to 0} x^p \cos\left(\frac{1}{x}\right) = 0 \]

5. **Conclusion**:
For the limit to equal \( f(0) = 0 \), we require \( p > 0 \).

### Final Condition:
The condition that should be imposed on \( p \) for \( f \) to be continuous at \( x = 0 \) is:
\[ p > 0 \]